Solve ∫ (x+1)(x-1)(x+2) wrt x

$$\int ( x + 1 ) ( x - 1 ) ( x + 2 ) d x$$

Evaluate

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}-2x+С$$

Solution Steps

Apply the distributive property by multiplying each term of x+1 by each term of x-1.

$$\int \left(x^{2}-x+x-1\right)\left(x+2\right)\mathrm{d}x$$

Combine -x and x to get 0.

$$\int \left(x^{2}-1\right)\left(x+2\right)\mathrm{d}x$$

Apply the distributive property by multiplying each term of x^{2}-1 by each term of x+2.

$$\int x^{3}+2x^{2}-x-2\mathrm{d}x$$

Integrate the sum term by term.

$$\int x^{3}\mathrm{d}x+\int 2x^{2}\mathrm{d}x+\int -x\mathrm{d}x+\int -2\mathrm{d}x$$

Factor out the constant in each of the terms.

$$\int x^{3}\mathrm{d}x+2\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{3}\mathrm{d}x with \frac{x^{4}}{4}.

$$\frac{x^{4}}{4}+2\int x^{2}\mathrm{d}x-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x^{2}\mathrm{d}x with \frac{x^{3}}{3}. Multiply 2 times \frac{x^{3}}{3}.

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}-\int x\mathrm{d}x+\int -2\mathrm{d}x$$

Since \int x^{k}\mathrm{d}x=\frac{x^{k+1}}{k+1} for k\neq -1, replace \int x\mathrm{d}x with \frac{x^{2}}{2}. Multiply -1 times \frac{x^{2}}{2}.

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}+\int -2\mathrm{d}x$$

Find the integral of -2 using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}-2x$$

If F\left(x\right) is an antiderivative of f\left(x\right), then the set of all antiderivatives of f\left(x\right) is given by F\left(x\right)+C. Therefore, add the constant of integration C\in \mathrm{R} to the result.

$$\frac{x^{4}}{4}+\frac{2x^{3}}{3}-\frac{x^{2}}{2}-2x+С$$

Differentiate w.r.t. x

$$\left(x+2\right)\left(x^{2}-1\right)$$